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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: probability 3
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[Post New]25 Mar 2008 21:31:41 IST
    Subject: probability 3
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sharmadon (2)

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6 couples have been seated along a long table,males on one side and females on the other.
 a. the probability that the birthdays of 6 males fall in exactly one calender month is 1/ 12^6
b. the probability that the birthdays of 6 males fall in exactly three calender months is 9900/ 12^6
c. if 4 persons are selected at random then the probability that 2 of them are male is 5/11
d. if 4 persons are selected at random then the probability that there is exactly one male and no two selected females are adjacent to each other is 17/165.
    
[Post New]25 Mar 2008 23:08:09 IST
    Subject: Re:probability 3
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sboosy (3063)

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\mbox{Each male could have his birthday on any one of the 12 months} \\ \\ \mbox{Thus denominator is} \ 12*12*12*12*12*12 = 12^6 \\ \\ \mbox{Now we can choose any one of the 12 months} \\ \\ \mbox{Thus the answer is} \ \frac{12}{12^6} = \frac{1}{12^5} \\ \\ \mbox{I think the answer given is wrong}
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[Post New]25 Mar 2008 23:17:30 IST
    Subject: Re:probability 3
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sboosy (3063)

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\mbox{2 males can be selected out of 6 males in} \ 6C_2 \ \mbox{ways} \\ \\ \mbox{2 females can be selected out of 6 females in} \ 6C_2 \ \mbox{ways} \\ \\ \mbox{Thus numerator is} \ 6C_2*6C_2 \\ \\ \mbox{The no. of ways of selecting 4 people out of 12 is} \ 12C_4 \\ \\ \mbox{Thus we have} \ \frac{6C_2*6C_2}{12C_4} = \frac{5}{11}
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